# I need help with a question of Foundations

1. Oct 26, 2004

### Paul Martin

I would like anyone to comment on a proposal I have written (it's short and intuitive) for a system of "Practical Numbers". The idea is to deny the Axiom of Choice, thus eliminating, or obviating, the notion of infinity and accepting the consequences of a largest integer and a smallest interval. I think all useful theorems of analysis will still hold in this system. The proposal can be found at http://paulandellen.com/essays/essay089.htm

My thanks to you in advance.

Paul Martin

2. Oct 26, 2004

### matt grime

The axiom of choice has nothing to do with whether or 'infinity exists' (an ill defined phrase, which we will take to mean there is a set whose cardinality is not finite, the existence of which in any model of ZF is guaranteed by the axiom of infinity, which would seem an 'infinitely' more obvious thing to deny if one didn't want an infinite set. Of course there is a difference between the axioms and a model of the axioms...)

Exactly what does 'deny' mean? Assume it is false? That still doesn't mean that the set of natural numbers has a largest element, as is rather trivial to see, even if one were to 'deny' the axiom of infinity in the sense of not include it in the axioms of ZF, that doesn't stop there being a model in which there is an infinite set. (ever heard of Skolem's Paradox? That's a beauty to frazzle your brain.)

None of the usual results in analysis will hold is my initial prediction since you've given no method to construct a model of the real numbers, what ever that might be, and of course in your system 1/n doesn't even tend to zero, so how it would be a useful place to do any analysis is a mystery.

Are you attempting to enter the annals of crankhood?

If I were you I'd go away and learn what the axiom of choice is before you start making false claims about it.

Last edited: Oct 26, 2004
3. Oct 27, 2004

### robert Ihnot

Matt grime seems pretty right on that. The Axiom of Choice has nothing to do with the set of all integers. If n is an integer, so in n+1, I don't see how you are going find any kind of upper bound there.

There have been serious mathematical studies which have denied the axiom of choice, indeed Cohen in 1963 showed that the Axiom of Choice is independent of Z-F set theory. The intuitionists have argued that all math must be constructed from the basics which is the set of integers.

As I was told in College years ago by a logic professor, nobody denies the existance of the set of all integers, after all, that is intuitive!

But, then again, maybe there is something in such an idea! I am tempted to start my own thread: Relativity and the counting process! https://www.physicsforums.com/showthread.php?p=354350#post354350

Last edited: Oct 27, 2004
4. Oct 27, 2004

### Hurkyl

Staff Emeritus
Furthermore, even if you took as an axiom that no infinite set exists, it does not follow that there are only finitely many integers (and thus a largest): it merely means that the integers don't form a set.

Last edited: Oct 27, 2004
5. Oct 27, 2004

### matt grime

There are reasons for assuming the axiom of choice, and for not doing so. WIthout it not every vector space has a basis, but with it we have the Banach Tarski Paradox. If we have the axiom of constructablity too (I think that's what I mean) then every set can be well ordered, and in particular there is a way to construct a theoretical game for two players where both have 'probability' one of winning. Of course, this is just theoretical and of no direct implication to the real world since you couldn't actually make such a game.

6. Oct 28, 2004

### Paul Martin

Hi Matt,

Thanks for your response. I will cave in for now and accept your statement that the Axiom of Choice has nothing to do with infinity. The issue that is important to me is that there are no infinities in mathematics. If denying (by that I mean excluding it from the ZF axioms) the Axiom of Infinity will do that, then that is what I would propose.

I suspect that your prediction that "none of the usual results of analysis will hold" was made without having read my proposal. If you had read it, you would have seen that Real Numbers are replaced by Practical Numbers. For computational purposes, Practical Numbers are every bit as useful as Real Numbers.

In my system, 1/n would tend to zero, but in the sense of a new and different definition of the limit which I sketched out in the essay. With new, but roughly equivalent, definitions for the limit, continuity, and convergence, I think all useful theorems (i.e. those applicable to real phenomena) of analysis could be proved in this system.

If you still don't think so after reading my proposal, please be specific as to why.

Paul Martin

7. Oct 28, 2004

### Paul Martin

Hi Robert,

You said, "If n is an integer, so in n+1, I don't see how you are going find any kind of upper bound there."

Here's how I see it: To even state the premise, "If n is an integer", you must first have the integers. Of course, if you do have the customary infinite set of integers, then I agree that n+1 is also an integer. But I want to back up and ask how do we get the integers in the first place?

You mentioned the set of integers as the "basics" of all mathematics. With my apologies to your professor, I am one person who denies the existence of the set of all integers. I say it is not intuitive.

In my view, mathematics deals only with concepts which are either taken as primitive, or they are defined, or they are derived. (And, of course, the definitions and the derivations must obey strict rules.)

It would be fair to take the infinite set of integers as primitive, and then you would have the familiar mathematics of real numbers. But, starting with Peano and his "successors", The integers are typically defined instead of taken as primitive I claim that you cannot legitimately define an infinite set of integers (or of anything else for that matter).

In your statement that I quoted, you implied that you can define the next integer after n simply by forming the sum n+1. But how is the sum itself defined? Addition is defined as an operation which is a function taking each pair of integers to another integer. In other words, you have to have the integers first before you can define the operation of addition.

But, no matter how you do the definition of the integers, in order to get an infinite set you have to assume that there is some automatic process that works blindingly fast and which has been working away for an exceedingly long time in order to produce it. Even if there conceptually were such a process, how could we be sure it has completed by now and actually produced the entire infinite set of integers?

There is nothing intuitive about this picture at all. It is preposterous. Yet, mathematicians accept it without much hesitation.

What seems more reasonable to me is to accept only definitions that have been explicitly made and written, either by human hands or by human made machines. So I'll readily accept the current approximate definition of Pi out to a trillion decimal places because it has actually been produced by a man-made machine. But I don't accept the existence of an approximation of Pi out to 10^trillion decimal digits. When it's finally computed out that far, I will.

By far the most common definitions of numbers today are found in the computer chips which embody algorithms to define and manipulate them. In each case, there is a practical limit on the size of the largest useful number. The usefulness of the larger of these numbers follows a pattern like the umbra and penumbra of a shadow as I described in my proposal (Thank you for reading and citing it, Robert.). Nothing prevents us from defining numbers as big as we need. But in all cases, the number of integers remains finite and the set of integers has an upper bound.

(There are actually two important upper bounds. In my proposal, the largest integer is equivalent to the largest integer for a given word size, which is the first of these bounds. The second bound, the largest Practical number, is the largest number which will not give an overflow when added to or multiplied by any smaller or equal number.)

When I expressed my doubts and concerns on this issue to my professors, I was told that the Axiom of Choice was what justified the assumption that an iterative inductive process like the definition of the succession of integers would ultimately produce an infinite set. It seemed like baloney to me then and it still seems like baloney to me now. If I'm wrong, I'd sure like someone to show me where and why.

It's fun talking to you, Robert.

Paul Martin

8. Oct 28, 2004

### Paul Martin

Hi Hurkyl,

Thanks for your response. Please see my reply to Robert for a discussion of how we get the integers.

Paul

9. Oct 28, 2004

### Hurkyl

Staff Emeritus
First, there are some subtle points that I think need addressed.

The first is the issue of denying an axiom -- this means much more than simply excluding the axiom from your theory. If I take ZFC, but omit the axiom of infinity, I'm left with a theory in which the set of natural numbers may or may not exist.

To deny the axiom, you have to take a new axiom that asserts the old axiom is false.

For your purposes, however, I think this may not even be sufficient -- you need an axiom that says, somehow, that all sets are finite.

The second is that of a class1 vs a set. It can be hard to distinguish the two concepts because the idea of a set is very closely modelled after the idea of a class, and ZFC was designed to be general enough that we'd never have to resort to using classes when doing everyday mathematics.

This is relevant to the current discussion -- while you can deny the existance of an infinite set, that does not mean there do not exist infinite classes. In particular, you can simply take the class of all sets!

1: I'm using the term "class" to refer to the objects that satisfy a logical proposition. For example, there is a class of "everything", because everything satisfies x = x.

10. Oct 28, 2004

### Hurkyl

Staff Emeritus
Next, I would like to suggest some topics you might find interesting. (and these topics are certainly not disjoint!)

You'll probably find studying at least some formal set theory worth your while. It would certainly help you develop your version of set theory, and it would probably be illuminating to see precisely how mathematics deals with these issues at the most fundamental level.

Next, I'd like to suggest formal logic, particularly first-order logic. IMHO, it just feels more constructive. It also has the particularly nice feature that everything in a theory that must be true has a proof!

I've found the basic theory of real closed fields (in particular, the fact the theory is complete) to be somewhat illuminating.

Nonstandard analysis is also quite interesting. Basically, it deals with theories A and B, where theory A is basically contained in theory B. However, if you limit your perspective properly, theories A and B will become identical, which allows you to do all sorts of interesting things. While I don't think the subject material will be relevant, per se, I think you might be able to extract some ideas and understanding from it.

11. Oct 28, 2004

### master_coda

But what's the value in restricting math in such a way? It's like arguing that computer scientists should only study computer programs that we can prove will halt or not halt, and should pretend that other programs do not exist until it has been proven that they halt (or do not).

Even if you feel that current concept of the integers is "absurd", that doesn't change the fact that the math and logic behind them is perfectly solid; this is all that mathematicians care about anyway.

If you want to study a specific number system, then do it; don't waste you time telling everyone that your system is better because it satisfies your personal prejudices about what is intuitive and what is not.

12. Oct 28, 2004

### Hurkyl

Staff Emeritus
Now, some practical stuff!

First off, have you seen Cantor's model of the natural numbers in set theory? Intuitively speaking, every Cantor natural number is defined to be the set of all smaller Cantor natural numbers. In particular:

0 := {} (the empty set)
1 := {0} = { {} } (the set containing the empty set)
2 := {0, 1} = { {}, {{}} }
3 := {0, 1, 2} = { {}, {{}}, { {{}, {{}} } }
...

In ZF, there's a class Ord of ordinal numbers which is defined by a logical proposition that says whether or not a set is of this form. However, since we only have finite sets in your set theory, the logical condition that defines this class will simply define the natural numbers.

So, even in this "finite set theory", we can define the class of natural numbers.

As an aside, the practical difference between a class and a set is that sets can be treated as objects, but classes cannot.

When dealing with set theory on this basic of a level, this model of the natural numbers is very convenient, because some of the basic operations are given by elementary set manipulation. Some examples:

The successor of N is given by N U {N} (U is union)
M <= N if and only if M is a subset of N
M < N if and only if M is an element of N
If Y is a set of (some) natural numbers, then the interesction of all the elements of Y is precisely the smallest element of Y.

As for functions, you are familiar with the set-theoretic definition. In particular, a function is supposed to be a set of ordered pairs. Well, you can also speak of a logic-theoretic definition of functions -- a (logic) function can be represented as a class of ordered pairs.

So, for example, I can speak of the function S which maps each natural numbers to its successor. It's defined by:

S(N, M) := M = N U {N}

But we would more conventionally write this as:

S(N) := N U {N}

even though the notation is the same, here I am denoting a logic function, not a set function.

Now, even though we cannot speak of the set of all integers, that's usually irrelevant. The nifty trick is to reduce our scope of inquiry to the relevant integers, but do it in an arbitrary way, so it's still valid for all integers!

That was a convoluted and impenetrable statement, wasn't it? Let me try via example.

Suppose we're still trying to define the relation <. We can give these two axioms:

N < S(N)
N < M and M < L implies N < L

(we really need more axioms to do this properly)

Now, I want to prove that 0 < M if M is any integer other than 0.

Normally, this would be a straightforward proof by induction:

0 < 1.
Suppose 0 < M. Then, 0 < M and M < S(M), therefore 0 < S(M).
Thus, 0 < M for all M >= 1.

But induction works by using the set of all integers, right? The usual proof of induction goes as follows:

Suppose there is a counterexample to the theorem. Then, there must be a least counterexample, X. However, there is a Y such that X = S(Y). Since the theorem is true for Y, it must also be true for X, which is a contradiction!

Actually, we don't need to work with the set of all integers to prove induction! We can use this slightly modified proof:

Theorem: Suppose P(0) is true and P(n) implies P( S(n) ). Then, P(n) for all natural numbers n.

Suppose the theorem is false. Then, there exists a natural number, M, such that P(M) is false.

Now, take the set {0, 1, 2, ..., M}. (This is a set, remember?) Now, among the elements of this set, there is a least counterexample! Call it X. X is not zero (because P(0) is true) Thus, there must be a Y such that X = S(Y). However, P(Y) is true (because Y < X, and X was the least counterexample), so P( S(Y) ) = P(X) must be true, which is a contradiction.

I've been sort of rambling on this, so I'll stop here and give things a chance to sink in.

13. Oct 28, 2004

### Hurkyl

Staff Emeritus
Well, I was slightly misleading... my example involving induction was entirely unnecessary -- given any class of natural numbers, you can always find the least element!

Recall the axiom of the subset: if I have some class of sets, C, then I can define the intersection of the entire class as follows:

Let a be some element of C.
Define I := {x in a | for all b: C(b) --> x in b }

By the axiom of subsets, I is, in fact, a set, and it is precisely the intersection of all of the sets in the class C. So, I didn't need to apply my little trick to prove that induction works, I can just use this.

It's still a useful trick, I just didn't select a good example.

14. Oct 29, 2004

### matt grime

You don't actually give any proper definitions to make it work. Your notion of penumbra is hazy (what is it for heaven's sake?) and requires the arbitrary removal of closure from the additive property of the natural numbers, and cartesian products (why should I be allowed to take the prodcut of two sets if I can't sum two numbers?) Why does Q even possess a square root?

nb Q is the 'largest integer' not the rationals.

Point 15 wrong. If every set is closed every set is open too, or would you need to redefine what a topology is?

Obviously all axioms are arbitrary ultimatelym but why must I forgo the fact that Z is a ring in order to satisfy your need to have "no infinities", especially since as we've seen nothing you've said forces there not to be some infinite object

You claim that you don't accept the set of integers if finite. Do you accept that there is at least 1 integer, say 1? Do you accept that 1+1 is also an integer, and that it is 2 and proceeding can we say 2+1 is an integer?

If the set is finite there is a maximal element, agreed? Say M, which is gotten by adding 1 M times, so why can't I logically add one more to M and get a bigger integer contradicting the assumption of finiteness.

Which of those steps is wrong in your opinion?

Note we do not assume that the whole set of integers exists and is infinite. We have 1 'existing' in some sense and we can add to it as we want. Purely a mathematical idea this nothing to do with what can be constructed (construct the number 1?) in any physical sense.

Why is it important to you that in your mathematics you have no infinities? They cause no problems.

And you can simply look at the statement of the axiom of choice and see that at no point does it "make an infinity exist" whatever the hell that might mean, and simply removing it from ZFC does not create a system that possesses a model that has only finitely many elements in it. The system without C is ZF and is well known indeed preferred by many for the philosophical issues this alleviates, but it still contains a copy of the natural numbers, or rather an inductive set.

It is perfectly possible to 'discretize' and choose your number system to be Z[e], by adjoining an free indeterminate to the integers. It's still an infinite set but all sums in it are required to be finite. Algebra abounds with examples of infinite systems like this where only finitely many terms are allowed (see the direct sum indexed by an infinite set for instance, as opposed to direct product).

Many mathematicians do not accept the axiom of choice (exactly what did cantor do with it that bothers you?).

If we accept its negation then there are many things that fail to be true since there are known to be many equivalent statements.

Infinite dimensional vector spaces wouldn't possess a basis springs to mind as the msot obvious one.

there are some good pieces of work by Conway et al that do set theory of infinite sets with out the axiom of choice.

Here's an older example (Bernstein-Schroeder)

If A and B are sets and there are injections A to B and B to A then there is a bijection between them.

Proof is axiom of choice free.

The dual statement involving surjections has a proof that requires the axiom of choice.

15. Oct 29, 2004

### Paul Martin

Hi Hurkyl,

Thanks for writing.

No, I don't think so. By "denying" the axiom I simply mean not including it with the axioms of the system.
No, I don't need that either. In the system I propose, we wouldn't ever introduce the terms "infinite" or "finite".We would accept only explicitly defined terms, including the numerals which are terms denoting specific numbers. Thus, the complete set of defined terms, in particular the complete set of defined numbers, would always be, to use a currently vernacular term, "finite". That would serve my purpose.
I agree that it is not relevant. I also agree that it might be interesting to pursue if I had the time and the inclination. Thanks for the suggestion all the same.

Paul

16. Oct 29, 2004

### Paul Martin

Yes, I have seen that and I am familiar with it. The part of it which is of interest to me is the ellipsis at the end. I would like you to give me a rigorous definition of what you mean by the ellipsis.
First of all, I'm not sure what you mean by "ZF". I'm guessing you mean the system of arithmetic as axiomatized by Zermelo and Fraenkel, but that's only a guess. Secondly, when you say "the natural numbers" I'm assuming you mean the customary infinite set of natural numbers. If so, please explain to me what "logical condition" will define the infinite set of natural numbers. That's what I think cannot be done rigorously. If we restrict ourselves to definitions which must be explicitly made, we must always have a system which we commonly refer to as "finite".

Thanks for the energy you put into this, Hurkyl.

Paul

17. Oct 29, 2004

### Paul Martin

Hi Master_Coda,

Since our world is grainy, it seems likely to me that a grainy math would be better suited to
describe the world than a smooth, continuous math would.
But the "facts" are otherwise. To be "perfectly solid", math must be consistent. Currently
accepted math and logic failed this test as soon as Cantor encountered his first paradox. I stand
with Kronecker and Brouwer on this issue. It was evidently exciting enough for mathematicians
to see the consequences of Cantor's work that they were willing to sweep the inconsistencies
under the rug. They euphemistically called the paradoxes and inconsistencies "antinomies".
Russell tried systematically to eliminate specific sets and propositions, with his Theory of Types,
in an attempt to isolate the inconsistencies. But, the way I see it, Kurt Goedel served up the fatal
blow. Goedel's theorem says that any system of math that includes the infinite set of integers
cannot be consistent and complete. In my opinion, that news should have caused mathematicians
to abandon the notion of infinity in mathematics for good. Instead, they continued to this day
with a system containing paradoxes and inconsistencies (er -- I mean "antinomies"). This is not at
all what I would call a "perfectly solid" system.
I agree. In fact, I think they should care more about it than they do. The inconsistencies don't
seem to bother them nearly as much as I think they should.

But I think my proposal holds more promise for Physicists than it does for Mathematicians. The
infinities that appear in the Physics equations are a real nuisance. In fact, they give Nobel Prizes
to Physicists who figure out ways of getting rid of some of the infinities. I would think that they
would find much more useful a mathematics that doesn't have infinities in it and which is grainy
just as the physical universe seems to be. Not only that, but none of their calculations would be
affected. In reality, all their calculations are already done in a system like the one I propose. (E.g.
no one has ever done a calculation using a value for Pi with an infinite number of digits.)

Paul

18. Oct 29, 2004

### Paul Martin

Hi Matt,

True. As I described it, my essay makes only an intuitive appeal.
That's a good one! LOL Not only my notion, but a penumbra itself is hazy. I'll give you the credit for intending the humor. I'll also assume your question was meant for readers (other than yourself, of course) who may not know what a penumbra is. If you know what a penumbra is, you may skip the next paragraph.

If you have a 1-foot diameter light source 10 feet from a wall and directly between them, 5 feet from the wall you have an opaque disc 1 foot in diameter, the disc will cast a shadow on the wall that has two distinct parts. In the middle will be a 1-foot diameter shadow which will be uniformly dark. If you sketch a diagram, you can see that no ray from the source can directly hit any part of this shadow. That part of the shadow is called the umbra. Surrounding the umbra is a more or less hazy shadow extending 1 foot beyond the umbra with the outside boundary making a 3 foot diameter circle concentric with the umbra. This band of shadow is called the penumbra. It is not uniform but instead is almost as dark as the umbra in regions near the umbra, and it is nearly as light as the unshaded wall near the outer boundary of the penumbra.

The metaphor of a shadow works well for explaining my proposal. If we assume the largest integer is Q, then you can think of a Cartesian plane with two concentric circles centered at the origin, one of radius Q the other of radius P = square root of Q. The outer ring is analogous to the panumbra and the inner circle to the umbra. The (x,y) pairs inside the inner circle represent all possible pairs of the Practical Numbers (those between -P and P), Sums and products are defined for all such pairs although a sum or product may not yield another practical number. But in all cases, the sum or product will be a number between -Q and Q. Numbers within the penumbra are called "impractical numbers" and may or may not have sums or products defined with other numbers.
It isn't arbitrary. It is necessary. If all sums and products are defined, then you have a situation where you can't define them all explicitly as I insist. If you assume that you can somehow acquire numbers without explicitly defining them, as is typically done, then you end up with infinite sets. It is that tacit acquisition assumption which I refuse to accept.

So, in my explicitly defined system, all numbers do not end up equal, even in their rights. Some have more or less operations defined on them than others. It's no different, except in scale, from what we have in typical number systems. There is an infinite (your concept, not mine) number of division operations which are not defined, viz. all those with zero divisors. My system simply has some sums, products, differences, and divisions which are also not defined. No problem.
Because it is a defined number in my system. It happens to be the largest number, but it is a number just the same. But you are probably getting at the situation where Q has an irrational square root in conventional mathematics. As is obvious, my system contains no irrational numbers, so the definition of 'square root' would have to involve a limiting process and assign the "closest" practical number to the desired root. This is exactly what we do in our computers and our calculations today. Nobody has ever worked with an irrational number. No problem.
I'm not sure what you are getting at here since I can't parse your statement.
I disagree with you here, but as I said in Point 15, "I will leave this for others to ponder."
I don't even know what "Z" is so I can't comment on its being a ring. My "need" is to accept terms that are explicitly and consistently defined. I am aware of no explicit and consistent definition of "infinity" or of any "infinite object". It isn't that I reject the notion of infinity. Instead I claim that the notion has never been properly introduced.
It goes wrong where you say, "contradicting the assumption of finiteness". If, by " logically add[ing] one more to M", you mean defining another number beyond M (your 'M' is my 'Q'), then, yes, you do get a bigger integer. You have defined a bigger system. (Now your M+1 is my
Q). But M+1 is still finite.

You are also wrong in characterizing the notion of "finiteness" as an assumption. I don't ever need to introduce that term into my system. The only reason you do, is to be able to make a distinction between my system and your notion of "infinity" which I claim you can't define in any consistent way. (NB the commonly accepted definition of infinity is Cantor's and he ran into inconsistencies with it right away.)
You have to be careful here if you want to be rigorous. Assuming that the whole "infinite" set of integers exists was the common foundation of arithmetic prior to Peano. He was the first to show how the integers could be constructed from an axiomatic base. But that "construction process" has holes in it which are my main objection. Similarly, we don't have 1 'existing' in any sense. To be rigorous, the number 1 must be defined. When I was studying this stuff in college 45 years ago, we spent six weeks defining the number 1 from a set-theoretic starting point. Bertrand Russell and Alfred North Whitehead took (if I remember right) a couple hundred pages at the beginning of their "Principia" to do the same. It's not trivial if you want to be rigorous.
I think it would work better for Physics as they try to describe our grainy universe.
In my view, they cause a host of problems, starting with Cantor's Paradox and followed by many others. They also cause horrendous problems in Physics calculations.

Thanks for all the time you have put in on this, Matt. I enjoy talking to you.

Paul

19. Oct 30, 2004

### matt grime

Firstly, mathematics isn't physical. The physical sciences use mathematics, but that infinite is well defined mathematically (contrary to your opinion) doesn't force infinite things into phyiscal reality. If it does it is either because of a flaw in the model, a flaw in the logic, or becuase there is indeed a singularity in the model there that needs to be physically interpreted. Just because the energy levels of a bound electron are quantized doesn't make the real number system invalid because you could divide the lowest energy by two and still have a real number that isn't the energy of a bound electron.

A set is infinite if it is not finite. A set is finite if it can be put in bijection with any of the inductive elements that arise in the inductive set in the axiom of infinity, or if you prefer, as we all should, with any of the sets {1,..,n} for some n in N.

nb N means the naturals, Q means the rationals informally in texts like this (usually we write $$\mathbb{Q}$$ hence why i qualified what Q is in my post incase anyone who hadn't read you essay thought it meant the rationals).

I'm sorry, what do you mean by Cantor's Paradox?

20. Oct 30, 2004

### Hurkyl

Staff Emeritus
Almost right; the Zermelo-Fraenkel axioms are for set theory, not arithmetic. (But, of course, arithmetic can be modeled in this set theory)

That is exactly wrong; much work went into developing a set theory that did not suffer from the inconsistencies. The most popular resulting theory is ZFC: the Zermelo-Fraenkel axioms + the axiom of choice.

All of the paradoxes of naive set theory arise from unrestricted comprehension: that is, saying "Let S be the set of all blah" where "blah" could be anything.

The usual solution is to only allow "safe" ways of making big sets: unions and power sets. Then use restriction (aka the axiom of subsets) to prove the existance of the set you really wanted.

First, allow me to define "transitive set":

A set S is transitive if and only if whenever x is an element of y and y is an element of S, then x is an element of S.

Then, the class Ord is defined to be all transitive sets that contain only transitive sets.

An example of a transitive set is:
{ {{0}}, {0}, 0 }
(where 0 is the empty set, as before). This is not an ordinal number, though, because it contains {{0}} which is not transitive (because it does not contain 0).

To be slightly more precise: the theory must be able to define a model of the integers, and the operations of multiplication and addition.

Whether they fit in a set or not is irrelevant to Godel's theorem.

And, it's not enough to simply contain the integers: the theory has to be able to define them too. See Tarski's completeness theorem -- Godel's theorem is not applicable to the first-order theory of real closed fields, even though every real closed field contains the integers.

(and, as a technical point, it also requires you to be working from a finite set of axioms)

Anyways, I don't see why you think this is a "fatal blow": while it is nice for a theory to be complete, it isn't required to do math. All this property means is that every statement either has a proof or disproof.